Section 10.1: Solutions of Differential Equations
An (ordinary) differential equation is an equation involving a
function and its derivatives.
That is, for functions P (x0 , x1 , . . . , xn ) and Q(x0 , . . . , xn ) of n + 1
variables, we say that the function f (t) (of one variable) satisfies
the differential equation
P (y, y 0 , . . . , y (n) ) = Q(f (t), . . . , f (n) (t))
if
P (f (t), f 0 (t), . . . , f (n) (t)) ≡ 0
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Examples
• The function f (t) = et satisfies the differential equation y 0 = y.
• The constant function g(t) ≡ 5 satisfies the differential
equation y 0 = 0.
• The functions h(t) = sin(t) and k(t) = cos(t) satisfy the
differential equation y 00 + y = 0.
• The function `(t) = ln(t) satisfies −(y 0 )2 = y 00 .
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Initial value problems
An initial value problem is a differential equation given together
with some requirements on the value of the function (or possibly
some of its derivatives) at certain points.
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Example
Find a function f (t) which satisfies f 0 (t) ≡ f (t) and f (0) = 30.
That is, solve the initial value problem y 0 = y and y(0) = 30.
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A solution
We know that if f (t) = Cet , for some constant C, then
f 0 (t) = Cet = f (t). So, such a function is a solution to the
differential equation y 0 = y. To solve the initial value problem we
need to specify C.
Evaluating,
30
= f (0) as f satisfies the initial value condition
= Ce0
= C
So, f (t) = 30et solves the initial value problem.
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Another example
Solve the initial value problem y 0 = cos(t) and y(0) = 1.
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Solution
If f 0 (t) = cos(t), then, by definition, f (t) is an antiderivative of
R
cos(t). We know cos(t)dt = sin(t) + C. Thus, f (t) = sin(t) + C
for some constant C. Evaluating at zero, we have
1 = f (0) = sin(0) + C = C. Thus, the solution to this initial value
problem is f (t) = sin(t) + 1.
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Constant solutions
In general, a solution to a differential equation is a function.
However, the function could be a constant function.
For example, all solutions to the equation y 0 = 0 are constant.
There are nontrivial differential equations which have some
constant solutions.
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Example
Find constant solutions to the differential equation
y 00 − (y 0 )2 + y 2 − y = 0
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Solution
y = c is a constant, then y 0 = 0 (and, a fortiori y 00 = 0). So, we
would need c2 − c = 0. That is, y ≡ 1 or y ≡ 0. Substituting, one
checks that these values satify the differential equation.
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Differential equations presented as narrative
problems
Newton’s Law of Cooling
Newton’s Law of Cooling states that the rate of cooling of an
object is proportional to the difference between its temperature and
the ambient temperature.
How are we to express this law in terms of differential equations?
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Solution
• Newton’s Law expresses a fact about the temperature of an
object over time. Express the temperature of the object at
time t as y(t).
• “rate of cooling” refers to a rate of change of the temperature,
ie to y 0 (t). The word “cooling” suggests that the derivative is
negative.
• To say that some quantity is proportional to another is to say
that there is some number α so that the first quantity is equal
to α times the second, independent of the time.
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Solution, continued
• In this formulation of the law, the ambient temperature is
taken to be fixed; call it A. The difference between the object’s
temperature and the ambient temperature is then y(t) − A.
• Putting these together, we obtain the differential equation
y 0 = α(y − A).
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Using Newton’s Law of Cooling
Suppose that an object initially having a temperature of 20◦ is
placed in a large temperature controlled room of 80◦ and one hour
later the object has a temperature of 35◦ . What will its
temperature be after three hours?
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Solution
We must solve the initial value problem y 0 = α(y − 80), y(0) = 20,
and y(1) = 35 where α is some unknown constant.
Clearly, the solution y ≡ 80 to the differential equation
y 0 = α(y − 80) does not satisfy the initial value conditions. So, we
may divide by y − 80, obtaining
y0
=α
y − 80
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Solution, continued
Integrating with respect to t from 0 to time T and substituting
u = y − 80 (so that du = y 0 dt) we have
Z T
αT =
αdt
0
Z
=
0
Z
T
y 0 dt
y − 80
y(T )−80
=
−60
du
u
y(T )−80
=
ln |u||−60
=
ln(60) − ln(80 − y(T ))
So ln(80 − y(T )) = ln(60) − αT .
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Solution, continued
Exponentiating both sides, we obtain
80 − y(T ) = 60e−αT
Evaluating at T = 1, we have
45 = 80 − 35 = 60e−α
So, α = ln( 34 ).
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Thus, y(T ) = 80 − 60eT ln( 4 ) = 80 − 60( 34 )T . So that
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y(3) = 80 − 60( 34 )3 = 80 − 60( 27
64 ) = 54 16 .
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Logistic equations
A logistic equation is a differential equation of the form
y 0 = αy(y − M ) for some constants α and M .
The logistic equation has the constant solutions y ≡ 0 and y ≡ M
and the nonconstant solution
y(t) =
M
−y(0) αM t
1 + ( My(0)
)e
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